# Question about ideals in quadratic extensions.

Let $\mathbb{L}/\mathbb{K}$ be a galois extension of number fields of degree 2 and classes number of $\mathbb{K}$ is $h(\mathbb{K})=1$. $O_{\mathbb{L}}, O_{\mathbb{K}}$ the rings of integers relatives to fields $\mathbb{L}$ and $\mathbb{K}$ respectively. I know that every ideal $I$ of $O_{\mathbb{L}}$ is generated by two elements $x, y$ such that $I=O_{\mathbb{L}}x+O_{\mathbb{L}}y$ and we can fix $x$ in the beginning, but with additional condition $h(\mathbb{K})=1$ I don't know if we have the result that $I$ is a free module of rank two over $O_{\mathbb{K}}$ and if any ideal $I$ is of the form $I=O_{\mathbb{K}}x+O_{\mathbb{K}}y$ for some $x, y\in O_{\mathbb{L}}$?

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## 1 Answer

Your $O_\mathbb{K}$ is a PID (principal ideal domain) because its class number is $1$. Over a PID any torsion free module is free, hence your $I$ is indeed of the form $I=O_{\mathbb{K}}x+O_{\mathbb{K}}y$.

More generally, without the class number condition, $I=O_{\mathbb{K}}x+Jy$ for some $x,y\in\mathbb{L}$ and an ideal $J\subseteq O_{\mathbb{K}}$ whose ideal class is uniquely determined by $I$. This is a special case of the structure theorem of torsion free modules over Dedekind domains.

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